Animated Solution for Mathematics - Differentiation: If the tangent to the curve, $y=e^x$ at a point $(c,e^c)$ and the normal to the parabola, $y^2=4x$ at the point $(1,2)$ intersect at the same point on the $x$-axis, then the value of $c$ is

Our first destination is the parabola defined by $y^2=4x$. We need to find the normal line at the point $(1,2)$.

To do this, we must first understand the slope of the tangent at that point. By differentiating $y^2=4x$ implicitly with respect to $x$, we obtain:

At the point $(1,2)$, the slope of the tangent $m_t$ is calculated as:

Since the normal is perpendicular to the tangent, its slope $m_n$ is the negative reciprocal of $1$, which is $m_n=-1$. Using the point-slope form $y-2=-1(x-1)$, we define the equation of the normal line.

To find where this line intersects the x-axis, we set $y=0$:

Thus, our normal line hits the x-axis at the point $(3,0)$.

Now, we turn our attention to the exponential curve $y=e^x$. We are looking for a tangent at the point $(c,e^c)$.

The derivative of $e^x$ is simply $e^x$, so the slope of the tangent at $x=c$ is $m_t=e^c$. Using the point-slope form, the equation of our tangent is:

Just as we did for the normal, we find the x-intercept by setting $y=0$:

Since $e^c$ is never zero, we can divide both sides by $e^c$, leaving us with $-1=x-c$, or $x=c-1$. The tangent hits the x-axis at the point $(c-1,0)$.

The problem states that these two lines intersect at the same point on the x-axis. This means the x-intercept of the tangent must equal the x-intercept of the normal.

We set the intercepts equal to each other:

Solving this linear equation, we find:

It is truly remarkable how two seemingly unrelated curves, governed by different mathematical laws, can be brought together by a single geometric condition. We have found our value, and in doing so, we have uncovered the hidden harmony between calculus and coordinate geometry.